The number of integers lying between 1000 and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We are looking for the number of $4$-digit integers $d_1d_2d_3d_4$ such that $d_1 + d_2 + d_3 + d_4 = 30$,where $1 \le d_1 \le 9$ and $0 \le d_2,

Vedclass · Accepted Answer

$84$

Vedclass · Answer

(A) We are looking for the number of $4$-digit integers $d_1d_2d_3d_4$ such that $d_1 + d_2 + d_3 + d_4 = 30$,where $1 \le d_1 \le 9$ and $0 \le d_2, d_3, d_4 \le 9$. Let $x_1 = d_1 - 1$,so $0 \le x_1 \le 8$. The equation becomes $(x_1 + 1) + d_2 + d_3 + d_4 = 30$,which simplifies to $x_1 + d_2 + d_3 + d_4 = 29$,with $0 \le x_1 \le 8$ and $0 \le d_2, d_3, d_4 \le 9$. Using the generating function method,we need the coefficient of $x^{29}$ in the expansion of $(1+x+...+x^8)(1+x+...+x^9)^3$. This is equivalent to the coefficient of $x^{29}$ in $(1-x^9)(1-x)^{-1} 	imes (1-x^{10})^3(1-x)^{-3} = (1-x^9)(1-3x^{10}+3x^{20}-x^{30})(1-x)^{-4}$. Expanding this,we look for terms that sum to $29$: $1 	imes \binom{29+4-1}{4-1} = \binom{32}{3} = 4960$ $-x^9 	imes \binom{20+4-1}{4-1} = -\binom{23}{3} = -1771$ $-3x^{10} 	imes \binom{19+4-1}{4-1} = -3 	imes \binom{22}{3} = -3 	imes 1540 = -4620$ $3x^{19} 	imes \binom{10+4-1}{4-1} = 3 	imes \binom{13}{3} = 3 	imes 286 = 858$ $3x^{20} 	imes \binom{9+4-1}{4-1} = 3 	imes \binom{12}{3} = 3 	imes 220 = 660$ $-3x^{29} 	imes \binom{0+4-1}{4-1} = -3 	imes 1 = -3$ Summing these: $4960 - 1771 - 4620 + 858 + 660 - 3 = 84$. Thus,the number of such integers is $84$.

The number of integers lying between $1000$ and $10000$ such that the sum of all the digits in each of those numbers becomes $30$ is:

Similar Questions

How many six-digit numbers are there in which no digit is repeated,even digits appear at even places,odd digits appear at odd places,and the number is divisible by $4$?

If $n$ is an integer with $0 \leq n \leq 11$,then the minimum value of $n!(11-n)!$ is attained when a value of $n$ equals to

The value of ${}^{47}C_4 + \sum_{j=1}^5 {}^{(52-j)}C_3$ is

There are $(n + 1)$ white and $(n + 1)$ black balls,with each set numbered $1$ to $n + 1$. The number of ways in which the balls can be arranged in a row so that no two adjacent balls are of the same colour is

The number of permutations of the letters $a_1, a_2, a_3, a_4, a_5$ in which the first letter $a_1$ does not occupy the first position and the second letter $a_2$ does not occupy the second position is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module